2
$\begingroup$

$f(x)$ is a differentiable function on the real line such that $\lim_{x \rightarrow \infty} f(x)=1$ and $\lim_{x \rightarrow \infty} f~'(x)=\alpha$. Then :

$(A) ~\alpha $ must be $0$

$(B) ~\alpha $ need not be $0$ but $|\alpha|<1$

$(C) ~\alpha > 1$

$(D) ~\alpha < -1$

Attempt:

$$f~'(x) = \lim_{h \rightarrow 0} \dfrac {f(x+h)-f(x)}{h}$$

$$\lim_{x \rightarrow \infty} f~'(x) =\lim_{x \rightarrow \infty} \lim_{h \rightarrow 0} \dfrac {f(x+h)-f(x)}{h}$$

$$\lim_{x \rightarrow \infty} f~'(x) = \lim_{h \rightarrow 0} \lim_{x \rightarrow \infty}\dfrac {f(x+h)-f(x)}{h}\tag{1}$$

$$\lim_{x \rightarrow \infty} f~'(x) = \lim_{h \rightarrow 0}\dfrac {1-1}{h} = 0 $$

Hence $\alpha$ must be equal to $0$.

Is step $(1)$ legal?

I mean, can we interchange the two limits?

When can we not interchange the two limits and when can we?

Thank you very much for your help in this regard.

$\endgroup$
  • 1
    $\begingroup$ The conclusion is correct: $\;\alpha =0\;$, yet step (1) needs justification and, in the general case, it can fail big time as you can see in some examples here: math.stackexchange.com/questions/15240/… $\endgroup$ – Timbuc Apr 29 '15 at 6:59
2
$\begingroup$

Justify each of the following steps:

We can try the following: for $\;n\in\Bbb N\;$ :

$$f(n+1)-f(n)=\frac{f(n+1)-f(n)}{(n+1)-n}=f'(c_n)\;\;,\;\;\;c_n\in (n,\,n+1)$$

using the MVT (Lagrange's Theorem) for differentiable functions.

From this it follows that $\;c_n\xrightarrow[n\to\infty]{}\infty\;$, and thus

$$\alpha=\lim_{n\to\infty}f'(c_n)=\lim_{n\to\infty}\left(f(n+1)-f(n)\right)=1-1=0$$

$\endgroup$
  • 1
    $\begingroup$ Got it. Thank you for your answer :) $\endgroup$ – MathMan Apr 29 '15 at 9:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.